help designing capacitors

[lokeycmos]

i cant afford to buy capacitors for my tesla coil, so im going to try my hand at making my own from sheets of glass and tinfoil. i have several sheets of 8.5"x11"glass. i only need it to be 8.6nF. what im wondering is if there are some formulas that will help me at least get a ball park figure based on area of the glass and area of the tin foil? any help would be appreciated! BTW, ive already tried the salt water bucket cap, it actually shattered while testing it.

 

[RCinFLA]

You are going to need the thickness of the glass. Look up the relative dielectric constant of glass. It is likely around 4.7 but can vary depending on glass make up. You also might want to be concerned about dielectric breakdown voltage on the glass, about 10 kV/mm thickness of the glass. Typical window glass will breakdown at about 30kV. Thicker glass has higher breakdown voltage but will yield less capacitance for given area.

https://en.wikipedia.org/wiki/Capacitance

 

[Dean Huster]

Remember that when using a capacitance equation, plate area means the area of your foil, not that of the glass plates.

To add to this problem, your plates must be a LOT smaller in dimension that the sheets of glass dielectric so that the cap won't arc around through the air from one side to the other. I've made single-sheet caps like this for a Tesla coil where the two plates were about 14 inches square with a two or three inch empty strip around the edge providing from four to six inches of air gap it would have to arc around.

For 8.6nF, you have some pretty small sheets. You can stack the plates and make a bigger cap with multiple plates that would look a bit like an old air-variable tuning cap, i.e., plate, glass, plate, glass, plate, glass, plate, glass, plate .... etc. You'll have to connect every other plate together so as you go along from one end of the cap to the other, it'll be AGBGAGBGAGBGAGB ... etc. where the A plates are connected to each other, the B plates are connected to each other and G represents the glass dielectric.

Add another problem. To get these parallel plate connections out to the edge of the glass where you can get to them, you've just halved that air gap which means that the plates will have to be even farther away from the edges in that dimension. If the cap is rectangular, the A plates will have to be connected together on the left side and the B plates on the right side, or you'll really reduce that air border.

Now, how will you attach a plate to glass so that it doesn't slither all over the place? Many adhesives will add a bit of thickness and lower the dielectric constant. I'd suggest something like 3M dry mount spray adhesive and go very lightly with it. Or, you could simply use "Scotch tape" to tape the corners in place.

Gets messy, doesn't it? But it would be fun to build something like that. Be ready for it to be pretty heavy and have a lot of volume to it.

The Wikipedia article is a little messy for our use. Here's the basic equation for calculating the value of your capacitance:


C = .224KA(n-1)/d

where C is the capacitance in pF, .224 is a constant for dealing with SAE measurements, K is the dielectric constant, A is the area of one plate in square inches and d is the distance between plates (thickness of the dielectric in most cases) in inches. A better equation for use with our modern calculators would be:

C = (.224E-12)KA(n-1)/d

where all is the same except that C will be in Farads rather than pF.

Now, we can rearrange that last equation to give us the number of plates required if we know everything else. It becomes

n = Cd/[(.224E-12)KA]

where dimensions are in inches and C is in Farads.

For your instance, let's assume that K = 4.7and C = 8.6E-9. If we give a 2-inch border around the edge of the glass, it means that our plate area will be 6.5 x 9 or 58.5 square inches. I'll assume the thickness of the glass to be 1/8 inch or 0.125 inch. Plugging things into the last equation, we'll end up with


n = 1 + [(8.6E-9)(0.125)] / [(0.2245E-12)(4.5)(58.5)] = about 19 plates, 18 sheets of glass.



Plugging the calculated 19 plates into our second equation, we'll have

C = [(0.224E-12)(4.7)(58.5)(19-1)] / 0.125 = 8.87pF, not too back considering all the slop involved.


I should add that the numerical constant 0.224 should be changed to 0.0885 if you want to use centimeters and square centimeters rather than inches for your calculations.

Send us a picture when you're done.